Suppose X has an exponential distribution with . Determine the following. Round the answers to 3 decimal places.

a) P(x<=0)
b) P(X>=2)
c) P(X<=1)
d) P(1<x<2)
e) Find the value of x such that P(X<x)=.05

To determine the probabilities in this case, we need to use the probability density function (PDF) of the exponential distribution. The PDF of an exponential distribution is given by:

f(x) = λ * e^(-λx), for x >= 0

Where:
- λ (lambda) is the rate parameter of the exponential distribution.

Given the rate parameter, we can then calculate the probabilities. Let's go through the calculations step by step:

a) P(x <= 0):

To calculate this probability, we need to integrate the PDF from 0 to infinity. However, since the exponential distribution is only defined for positive values of x, P(x <= 0) is equal to 0.

b) P(X >= 2):

Here, we need to calculate the complement of the probability, which means finding P(X < 2) and subtracting it from 1. To calculate P(X < 2), we integrate the PDF from 0 to 2. Let's denote the integral as I(0, 2):

I(0, 2) = ∫[0 to 2] λ * e^(-λx) dx

To calculate this integral, we can use the following indefinite integral:

∫ e^(-λx) dx = (-1/λ) * e^(-λx) + C

Applying the limits of integration, we have:

I(0, 2) = (-1/λ) * e^(-λx) from 0 to 2

To find P(X >= 2), we subtract I(0, 2) from 1:

P(X >= 2) = 1 - I(0, 2)

c) P(X <= 1):

To calculate this probability, we integrate the PDF from 0 to 1:

I(0, 1) = ∫[0 to 1] λ * e^(-λx) dx

Similarly to the previous calculation, we use the indefinite integral:

∫ e^(-λx) dx = (-1/λ) * e^(-λx) + C

Applying the limits of integration, we get:

I(0, 1) = (-1/λ) * e^(-λx) from 0 to 1

d) P(1 < x < 2):

To calculate this probability, we need to find the difference between P(X < 2) and P(X < 1):

P(1 < x < 2) = P(X < 2) - P(X < 1)

e) Find the value of x such that P(X < x) = 0.05:

To find the value of x, we need to find the inverse of the cumulative distribution function (CDF) of the exponential distribution. The CDF of the exponential distribution is given by:

F(x) = 1 - e^(-λx)

We set F(x) equal to 0.05 and solve for x:

0.05 = 1 - e^(-λx)

e^(-λx) = 0.95

Taking the natural logarithm (ln) of both sides:

-ln(0.95) = -λx

Solving for x:

x = -ln(0.95) / λ

This value of x gives P(X < x) = 0.05.

Note: In order to provide specific numerical answers, we need to know the value of λ (the rate parameter).